Exercise on Rudin about $R^2$ measurable

330 Views Asked by At

I'm thinking about exercise 9 on Rudin's Real and Complex Analysis chapter 8:

$E$ is dense in $\mathbb{R}^1$ and $f$ is a real function on $\mathbb{R}^2$ such that:

(a) $f_x$ is Lebesgue measurable for each $x\in E$;

(b) $f^y$ is continuous for almost all $y\in \mathbb{R}^1$.

Prove that $f$ is $\mathbb{R}^2$ Lebesgue measurable.

My idea is to construct a sequence $\{f_n\}_n$ to approach $f$ just like the exercise above, but fail. Who can give some hints for this problem? Any suggestion is appreciated.

Thank all of you!

1

There are 1 best solutions below

0
On

Take these hints one at a time because reading them all practically gives it away.

Hint 1: Since $f^y$ is continuous for almost all $y \in R$, then for $Y = \{ y : f^y \text{ continuous} \}$, $m(R - Y) = 0$. Define $g = f \vert_{R \times Y}$, and its easy to see that $m(f > a) = m(g > a)$ since $m(R - Y) = 0$. So, we now restrict ourselves to $g$.

Hint 2: The point-wise limit of Lebesgue mbl. functions is Lebesgue mbl.

Still lost? The last one will help you build functions $g^{(\epsilon)}$ whose point-wise limit is $g$.

Hint 3: For $\epsilon > 0$, define the half open interval $I_z = [z \epsilon, (z+1) \epsilon)$. These half open intervals cover $R$. By the density of $E$, we know there is a point $x_z \in E \cap I_z$. Now generate a function $g^{(\epsilon)}$ whose value for (x, y) is dependent on $(x_z, y)$. Prove $g^{(\epsilon)}$ is Lebesgue mbl.